Answer:
Check the solution below
Step-by-step explanation:
2) Given the equation
x +y =5... 1 and
x-y =3 ... 2
Add both equations
x+x = 5+3
2x = 8
x = 8/2
x = 4
Substitute x = 4 into 1:
From 1: x+y = 5
4+y= 5
y = 5-4
y = 1
3) Given
x+3y =15 ... 1
2x+7y=19 .... 2
From 2: x = 15-3y
Substitute into 2
2(15-3y)+7y = 19
30-6y+7y = 19
30+y = 19
y = 19-30
y = -11
Substitute y=-11 into x = 15-3y
x =15-3(-11)
x = 15+33
x = 48
The solution set is (48, -11)
4) given
x/2 +y/3 =0 and x+2y=1
From 1
(3x+2y)/6 = 0
3x+2y = 0.. 3
x+2y= 1... 4
From 4: x = 1-2y
Substutute
3(1-2y) +2y = 0
3-6y+2y = 0
3 -4y = 0
4y = 3
y = 3/4
Since x = 1-2y
x = 1-2(3/4)
x = 1-3/2
x= -1/2
The solution set is (-1/2, 3/4)
5) Given
5.x=1/2 and y =x +1 then solution is
We already know the vkue of x
Get y
y= x+1
y = 1/2 + 1
y = 3/2
Hence the solution set is (1/2, 3/2)
6) Given
3x +y =5 and x -3y =5
From 3; x = 5+3y
Substitute into 1;
3(5+3y)+y = 5
15+9y+y = 5
10y = 5-15
10y =-10
y = -1
Get x;
x = 5+3y
x = 5+3(-1)
x = 5-3
x = 2
Hence two solution set is (2,-1)
There is a lot of numbers
Answer:
14
Step-by-step explanation:
its an formula which comes in algebraic expression
<h3>
Answer: Choice D. 
</h3>
The red angle markers show those two angles are congruent. That's one "A" of "ASA". The S refers to a congruent pair of sides. We don't have any tickmarks to indicate congruent pairs; however, we do know that QR = QR is a shared side that overlaps (reflexive theorem). So this is the "S" in "ASA".
The thing missing is the angle Q of the top triangle, and also of the bottom triangle as well. If we know those two angles are congruent, then we have enough info to use ASA. More specifically, if we know that , then we can use ASA.
One thing to notice is that the other answer choices involve side lengths and not angles. This implies that if A, B or C were one of the answers, then we would have something like SAS or SSS. But instead we want ASA. So we can immediately rule choices A,B, and C out.
Answer:
B) ∠A and ∠B
Step-by-step explanation:
AB is the line segment formed with endpoints at A and B. This means it lies between the angle with vertex at A, ∠A, and the angle with vertex at B, ∠B.
